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Saturday, 18 March 2017

SBI PO Quadratic Equation Questions and Answers Set – 7

     
     
     
    1 . Directions (Q. 1-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
    (1) if x > y
    (2) if x $\leq$y
    (3) if x < y
    (4) if x $\leq$
    5) if x = y or the relationship between x and y cannot be established.

    $Q.$
    I. $x^ 2 $ + 12x + 36 = 0
    II. $y^ 2 $ + 15y + 56 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    2 . I. $x^ 2$ = 35
    II. $y^ 2 $ + 13y + 42 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    3 . I. $2x^ 2 $ - 3x - 35 = 0
    II. $y^ 2 $ - 7y + 6 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    4 . I. $6x^ 2 $ - 29x + 35 = 0
    II. $2y^ 2 $ - 19y + 35 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    5 . I.$12x^ 2 $ - 47x + 40 = 0
    II. $4y^ 2$ + 3y - 10 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    6 . I. $x^ 2$ + 3x - 28 = 0
    II.$ y^ 2 $ - 11y + 28 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    7 . I. $6x^ 2 $ - 17x + 12 = 0
    II.$ 6y^ 2 $ - 7y + 2 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    8 . I. x = $\sqrt{256}\over \sqrt{576}$
    II. $3y^ 2$ + y-2 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    9 . I. $x^ 2$ = 64
    II. y$^ 2$ = 9y

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x \leq y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    10 . I. $x^ 2$ + 6x - 7 = 0
    II. 41y + 17 = 140

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
      Answers & Solutions
       
      1 .    
      Answer : Option A
      Explanation :
      I. $x^ 2$ + 12x + 36 = 0
      or, $(x + 6)^ 2$ = 0
      or, x + 6 = 0
      or, x = - 6

      II. $y^ 2$ + 15y + 56 = 0
      or, $y^ 2$ + 7y + 8y + 56 = 0
      or, y(y + 7) + 8(y + 7) = 0
      or, (y + 7) (y + 8) = 0
      y = -7, -8

      x > y
      2 .    
      Answer : Option A
      Explanation :
      I. $x^ 2$ = 35
      $x^ 2$ = $\sqrt{35}$

      II. $ y 2$ + 13y + 42 = 0
      or, $ y 2$ + 6y + 7y + 42 = 0
      or, y(y + 6) + 7(y + 6) = 0
      or, (y + 6) (y + 7) = 0
      y = -6, - 7

      x > y
      3 .    
      Answer : Option D
      Explanation :
      I. $2x^ 2$ - 3x - 35 = 0
      or, $2x^ 2 $ - 10x + 7x - 35 = 0
      or, 2x(x - 5) + 7(x - 5) = 0
      or, (2x + 7) (x - 5) = 0
      x = -$7\over 2$ , 5

      II. $y^ 2$ - 7y + 6 = 0
      or, $ y^ 2$ - y - 6y + 6 = 0
      or, y(y - 1) - 6(y - 1)
      or, (y - 1)(y - 6) = 0
      y = 1, 6

      No relation can be established between x and y.
      4 .    
      Answer : Option D
      Explanation :
      I. $6x^ 2$ - 29x + 35 = 0
      or, $ 6x^ 2$ - 15x - 14x + 35 = 0
      or, 3x(2x - 5) -7(2x - 5) = 0
      or, (3x - 7) (2x - 5) = 0
      x = $7\over 3$, $5\over 2$

      II. $2y^ 2$ - 19y + 35 = 0
      or, $2y^ 2 $ - 14y - 5y + 35 = 0
      or, 2y(y - 7) -5 (y - 7) = 0
      or, (2y - 5)(y - 7) = 0
      y = $5\over 2$, 7

      X $\leq$ y
      5 .    
      Answer : Option B
      Explanation :
      I. $ 12x^ 2$ - 47x + 40 = 0
      or, $ 12x^ 2$ - 32x - 15x + 40 = 0
      or, 4x(3x - 8) -5(3x - 8) = 0
      or, (4x - 5) (3x - 8) = 0
      x = $5\over 4$, $8\over 3$

      II. $4y^ 2$ + 3y - 10 = 0
      or, $ 4y^ 2$ + 8y - 5y - 10 = 0
      or, 4y(y + 2) -5(y + 2) = 0
      or, (4y - 5) (y + 2) = 0
      y = $5\over 4$, - 2
      x $\geq$ y
      6 .    
      Answer : Option D
      Explanation :
      I. $x^ 2$ + 7x - 4x - 28 = 0
      or, x(x + 7) - 4 (x + 7) = 0
      or, (x - 4)(x + 7) = 0
      x = 4, - 7

      II. $y^ 2 $ - 11y + 28 = 0
      or, $y^ 2$ - 7y - 4y + 28 = 0
      or, y (y - 7) -4(y - 7) = 0
      or, (y - 4) (y - 7) = 0
      y = 4, 7

      x $\leq$ y
      7 .    
      Answer : Option A
      Explanation :
      I. $6x^ 2 $ - 17x + 12 = 0
      or, $6x^ 2$ - 9x - 8x + 12 = 0
      or, 3x (2x - 3) - 4 (2x - 3) = 0
      or, (3x - 4) (2x - 3) = 0
      x = $4\over 3$, $3\over 2$

      II. $6y^ 2$ - 3y - 4y + 2 = 0
      or, 3y (2y - 1) - 2 (2y - 1) =0
      or, (3y - 2) (2y - 1) = 0
      y = $2\over 3$, $1\over 2$

      x > y
      8 .    
      Answer : Option B
      Explanation :
      I. x = $\sqrt{256}\over \sqrt{576}$
      x = $16\over 24$ = $2\over 3$
      II. $3y^ 2$ + y - 2 = 0
      or, $3y^ 2$ + 3y - 2y - 2 = 0
      or, 3y (y + 1) - 2(y + 1) = 0
      or, (3y - 2) (y + 1) = 0
      y = $2\over 3$, -1

      $x \geq y$
      9 .    
      Answer : Option D
      Explanation :
      I. $x^ 2$ = 64
      x = ± 8

      II.$y^ 2$ = 9y
      or, $y^ 2$ - 9y = 0
      or, y (y - 9) = 0
      y = 0, 9

      no relationship can be established between x and y
      10 .    
      Answer : Option C
      Explanation :
      I. $x^ 2$ + 6x - 7 = 0
      or, $ x^ 2$ + 7x - x - 7 = 0
      or, x(x + 7) -1 (x + 7)= 0
      or, (x - 1) (x + 7) = 0
      x = 1, -7

      II. 41y + 17 = 140
      or, 41y = 140 - 17 = 123
      y = $123\over 41$ = 3

      x < y

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